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Dafna11 [192]
3 years ago
15

What are the x-intercepts of the graph of the function below? y = x2 + 2x-15

Mathematics
2 answers:
Maurinko [17]3 years ago
7 0

Answer:

x = - 5, x = 3

Step-by-step explanation:

To find the x- intercepts let y = 0, that is

x² + 2x - 15 = 0

Consider the factors of the constant term (- 15) which sum to give the coefficient of the x- term (+ 2)

The factors are + 5 and - 3, since

5 × - 3 = - 15 and 5 - 3 = + 2, hence

(x + 5)(x - 3) = 0

Equate each factor to zero and solve for x

x + 5 = 0 ⇒ x = - 5

x - 3 = 0 ⇒ x = 3

x- intercepts at (- 5, 0) and (3, 0)

timama [110]3 years ago
6 0

Answer:

-15

Step-by-step explanation:

the y-intercept would be -15

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Rather than carrying out IBP several times, let's establish a more general result. Let

I(n)=\displaystyle\int x^ne^x\,\mathrm dx

One round of IBP, setting

u=x^n\implies\mathrm du=nx^{n-1}\,\mathrm dx

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\displaystyle I(n)=x^ne^x-n\int x^{n-1}e^x\,\mathrm dx

I(n)=x^ne^x-nI(n-1)

This is called a power-reduction formula. We could try solving for I(n) explicitly, but no need. n=5 is small enough to just expand I(5) as much as we need to.

I(5)=x^5e^x-5I(4)

I(5)=x^5e^x-5(x^4e^x-4I(3))=(x-5)x^4e^x+20I(3)

I(5)=(x-5)x^4e^x+20(x^3e^x-3I(2))=(x^2-5x+20)x^3e^x-60I(2)

I(5)=(x^2-5x+20)x^3e^x-60(x^2e^x-2I(1))=(x^3-5x^2+20x-60)x^2e^x+120I(1)

I(5)=(x^3-5x^2+20x-60)x^2e^x+120(xe^x-I(0))

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I(0)=\displaystyle\int e^x\,\mathrm dx=e^x+C

so we end up with

I(5)=(x^4-5x^3+20x^2-60x+120)xe^x-120e^x+C

I(5)=(x^5-5x^4+20x^3-60x^2+120x-120)e^x+C

and the antiderivative is

\displaystyle\int2x^5e^x\,\mathrm dx=(2x^5-10x^4+40x^3-120x^2+240x-240)e^x+C

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3x^2+18x-21

Let's start by taking 3 out of the equation.

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Now, when we factor, we need to find two numbers that add up to 6 and multiply to -7.

The numbers would be 7 and -1.

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